Vorlesung Datenstrukturen und Algorithmen Letzte Vorlesung 2018 - - PowerPoint PPT Presentation

Vorlesung Datenstrukturen und Algorithmen Letzte Vorlesung 2018

Felix Friedrich, 30.5.2018 Map/Reduce Sorting Networks Prüfung

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MAP AND REDUCE AND MAP/REDUCE

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Summing a Vector

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+ + + + + + + Accumulator + + + + + + + + Divide and conquer

Q: Why is the result the same? A: associativity: (a+b) + c = a + (b+c)

Summing a Vector

+ + + + + + + + Divide and conquer + + + + Is this correct? + + +

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Only if the operation is commutative: a + b = b + a

Reductions

Simple examples: sum, max Reductions over programmer-defined operations

– operation properties (associativity / commutativity) define the correct

executions

– supported in most parallel languages / frameworks – powerful construct

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C++ Reduction

• std::accumulate (requires associativity)
• std::reduce (requires commutativity, from C++17, can specify execution

policy)

std::vector<double> v; ... double result = std::accumulate( v.begin(), v.end(), 0.0, [](double a, double b){return a + b;} );

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Elementwise Multiplication

Map

7 x

+ Multiply

x x x x x x x

Scalar Product

Map Reduce

8 x

+ Multiply

x x x x x x x

+ + + + + + + + Accumulate

C++ Scalar Product (map + reduce)

// example data

std::vector<double> v1(1024,0.5); auto v2 = v1; std::vector<double> result(1024); // map std::transform(v1.begin(), v1.end(), v2.begin(), result.begin(), [](double a, double b){return a*b;}); // reduce double value = std::accumulate(result.begin(), result.end(), 0.0); // = 256

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Map & Reduce = MapReduce

Combination of two parallelisation patterns result = 𝑔 in1 ⊕ 𝑔 in2 ⊕ 𝑔(in3) ⊕ 𝑔(in4) 𝑔 = map ⊕ = reduce (associative) Examples: numerical reduction, word count in document, (word, document) list, maximal temperature per month over 50 years (etc.)

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Motivating Example

Maximal Temperature per Month for 50 years

• Input: 50 * 365 Days / Temperature pairs
• Output: 12 Months / Max Temperature pairs

Assume we (you and me) had to do this together. How would we distribute the work? What is the generic model? How would we be ideally prepared for different reductions (min, max, avg)?

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Maximal Temperature per Month: Map

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data 01 / 5 01 / 8 01 / 8 .... 03 / 12 03 / 14 .... 05/20 05/19 05/20 ... 07/28 07/38 ... ... ... ... ... ... ... 03/14 03/18 .... ...

each map-process gets day/temperature pairs and maps them to month/temperature pairs

Maximal Temperature per Month: Shuffle

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01 / -5 01 / -8 01 / -8 .... 03 / 12 03 / 14 .... 05/20 05/19 05/20 ... 07/28 07/38 ... ... ... ... ... ... ... 03/14 03/18 .... Jan 01 / -5 01 / -8 01 / -8 .... Feb 02 / 13 02 / 14 02 / 12 .... Mar 03 / 13 03 / 14 03 / 12 .... April 04 / 23 04 / 24 04 / 22 .... Dec 12 / 0 12 / -2 12/ 2 .... ...

data gets sorted / shuffled by month

Maximal Temperature per Month: Reduce

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Jan 01 / -5 01 / -8 01 / -8 .... Feb 02 / 13 02 / 14 02 / 12 .... Mar 03 / 13 03 / 14 03 / 12 .... April 04 / 23 04 / 24 04 / 22 .... Dec 12 / 0 12 / -2 12/ 2 .... Jan 18 Feb 20 Mar 22 April 28 Dec 20

each reduce process gets its own month with values and applies the reduction (here: max value) to it

Map/Reduce

A strategy for implementing parallel algorithms.

• map: A master worker takes the problem input, divides it

into smaller sub-problems, and distributes the sub-problems to workers (threads).

• reduce: The master worker collects sub-solutions from the

workers and combines them in some way to produce the

• verall answer.

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Map/Reduce

Frameworks and tools have been written to perform map/reduce.

• MapReduce framework by Google
• Hadoop framework by Yahoo!
• related to the ideas of

Big Data and Cloud Computing

• also related to functional programming

(and actually not that new) and available with the Streams concept in Java (>=8)

• Map and reduce are user-supplied plug-ins, the rest is provided by the

frameworks.

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Jeffrey Dean and Sanjay Ghemawat. 2008. MapReduce: simplified data processing on large clusters. Commun. ACM 51, 1 (January 2008), 107-113. DOI=10.1145/1327452.1327492 http://doi.acm.org/10.1145/1327452.1327492

MapReduce on Clusters

You may have heard of Google’s “map/reduce” or Amazon's Hadoop Idea: Perform maps/reduces on data using many machines

• The system takes care of distributing the data and managing fault tolerance
• You just write code to map one element (key-value part) and reduce elements

(key-value pairs) to a combined result

Separates how to do recursive divide-and-conquer from what computation to perform

• Old idea in higher-order functional programming transferred to large-scale

distributed computing

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Example

Count word occurences in a very large file File =

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how are you today do you like the weather outside I do I do I wish you the very best for your exams. GBytes

Mappers

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huge file part 1 part 2 part 3 key/value pairs (e.g. position / string) <0, "how are you today"> <15, "do you like ..."> <35, "I do"> <39, "I do"> <43, "I wish you the very best"> <70, "for your exams"> mapper 1 mapper 2 mapper 3

DISTRIBUTED

Mappers

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key/value pairs (e.g. position / string) <0, "how are you today"> <15, "do you like ..."> <35, "I do"> <39, "I do"> <43, "I wish you the very best"> <70, "for your exams"> mapper 1 mapper 2 mapper 3 key/value pairs (word, count)

<"how",1> <"are",1> <"you",1> ... <"I",1> <"do",1> <"I",1> <"do",1> <"I",1> <"wish",1> <"you",1> ...

input

• utput

Shuffle / Sort

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mapper 1 mapper 2 mapper 3 unique key/value pairs (shuffled) (word, counts)

<"how",1> <"are",1> <"you",1> ... <"I",1> <"do",1> <"I",1> <"do",1> <"I",1> <"wish",1> <"you",1> ... <"do",1,1,1> <"for",1> ....

reducer 1 reducer 2

<"are",1> <"best",1> ...

key/value pairs (word, count)

DISTRIBUTED

Reduce

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unique key/value pairs (shuffled) (word, counts)

<"do",1,1,1> <"for",1> ....

reducer 1 reducer 2

<"are",1> <"best",1> ...

target file(s)

are 1 best 1 you 3 ... do 3 for 1 I 3 ...

input

• utput

SORTING NETWORKS

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Lower bound on sorting

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Simple algorithms: O(n2) Fancier algorithms: O(n log n) Comparison lower bound: (n log n) Insertion sort Selection sort Bubble Sort Shell sort … Heap sort Merge sort Quick sort (avg) … Horrible algorithms: Ω(n2) Bogo Sort Stooge Sort

Comparator

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x y min(x,y) max(x,y)

<

x y min(x,y) max(x,y)

shorter notation:

void compare(int&a, int&b, boolean dir) { if (dir==(a,b)){ std::swap(a,b); } }

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a b a b

> <

Sorting Networks

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1 5 4 3 5 1 4 3 1 3 4 5 3 4 3 1 4 5

Sorting Networks are Oblivious (and Redundant)

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2:3 4:3 2:1 4:1 2:4 3:4 2:1 3:1 1:3 4:3 1:2 4:2 1:4 3:4 1:2 3:2 2:4 2:4 2:3 2:3 1:4 1:4 1:3 1:3 1:3 1:4 2:3 2:4 3:4 3:4 1:2 𝑦1 𝑦2 𝑦3 𝑦4 Oblivious comparison tree redundant cases

Recursive Construction : Insertion

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𝑦1 𝑦2 𝑦3 𝑦𝑜−1 𝑦𝑜 𝑦𝑜+1 sorting network . . . . . . . . .

Recursive Construction: Selection

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𝑦1 𝑦2 𝑦3 𝑦𝑜−1 𝑦𝑜 𝑦𝑜+1 sorting network . . . . . . . . .

Applied recursively..

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insertion sort bubble sort with parallelism: insertion sort = bubble sort !

Question

How many steps does a computer with infinite number of processors (comparators) require in order to sort using parallel bubble sort? Answer: 2n – 3 Can this be improved ? How many comparisons ? Answer: (n-1) n/2 How many comparators are required (at a time)? Answer: n/2 Reusable comparators: n-1

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Improving parallel Bubble Sort

Odd-Even Transposition Sort: 9 8 2 7 3 1 5 6 4 1 8 9 2 7 1 3 5 6 4 2 8 2 9 1 7 3 5 4 6 3 2 8 1 9 3 7 4 5 6 4 2 1 8 3 9 4 7 5 6 5 1 2 3 8 4 9 5 7 6 6 1 2 3 4 8 5 9 6 7 7 1 2 3 4 5 8 6 9 7 8 1 2 3 4 5 6 8 7 9 1 2 3 4 5 6 7 8 9

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void oddEvenTranspositionSort(std::vector<T>& v, boolean dir) { for (int i = 0; i<v.size(); ++i){ for (int j = i % 2; j+1<n; j+=2) compare(v[i],v[j],dir); } }

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Improvement?

Same number of comparators (at a time) Same number of comparisons But less parallel steps (depth): n

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In a massively parallel setup, bubble sort is thus not too bad. But it can go better...

Parallel sorting

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Prove that the two networks above sort four numbers. Easy?

depth = 4 depth = 3

Zero-one-principle

Theorem: If a network with 𝑜 input lines sorts all 2𝑜 sequences of 0s and 1s into non-decreasing order, it will sort any arbitrary sequence

• f 𝑜 numbers in nondecreasing order.

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Proof

Assume a monotonic function 𝑔(𝑦) with 𝑔 𝑦 ≤ 𝑔(𝑧) whenever 𝑦 ≤ 𝑧 and a network 𝑂 that sorts. Let N transform (𝑦1, 𝑦2, … , 𝑦𝑜) into (𝑧1, 𝑧2, … , 𝑧𝑜), then it also transforms (𝑔(𝑦1), 𝑔(𝑦2), … , 𝑔(𝑦𝑜)) into (𝑔(𝑧1), 𝑔(𝑧2), … , 𝑔(𝑧𝑜)). Assume 𝑧𝑗 > 𝑧𝑗+1for some 𝑗, then consider the monotonic function 𝑔(𝑦) = ቊ0, 𝑗𝑔 𝑦 < 𝑧𝑗 1, 𝑗𝑔 𝑦 ≥ 𝑧𝑗 N converts (𝑔(𝑦1), 𝑔(𝑦2), … , 𝑔(𝑦𝑜)) into 𝑔 𝑧1 , 𝑔(𝑧2 , … 𝑔 𝑧𝑗 , 𝑔 𝑧𝑗+1 , … , 𝑔(𝑧𝑜))

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𝑦 not sorted by 𝑂 ⇒ there is an 𝑔 𝑦 ∈ 0,1 𝑜 not sorted by N ⇔ 𝑔 sorted by N for all 𝑔 ∈ 0,1 𝑜 ⇒ 𝑦 sorted by N for all x

Argue: If x is sorted by a network N then also any monotonic function of x. 20 8 1 30 5 9 8 5 1 9 20 30 10 4

• 1

99 2 9 4 2

• 1

9 10 99 Show: If x is not sorted by the network, then there is a monotonic function f that maps x to 0s and 1s and f(x) is not sorted by the network 20 8 1 30 5 9 9 5 1 8 20 30 1 1 0 1 1 1 1

Proof

Assume a monotonic function 𝑔(𝑦) with 𝑔 𝑦 ≤ 𝑔(𝑧) whenever 𝑦 ≤ 𝑧 and a network 𝑂 that sorts. Let N transform (𝑦1, 𝑦2, … , 𝑦𝑜) into (𝑧1, 𝑧2, … , 𝑧𝑜), then it also transforms (𝑔(𝑦1), 𝑔(𝑦2), … , 𝑔(𝑦𝑜)) into (𝑔(𝑧1), 𝑔(𝑧2), … , 𝑔(𝑧𝑜)). Assume 𝑧𝑗 > 𝑧𝑗+1for some 𝑗, then consider the monotonic function 𝑔(𝑦) = ቊ0, 𝑗𝑔 𝑦 < 𝑧𝑗 1, 𝑗𝑔 𝑦 ≥ 𝑧𝑗 N converts (𝑔(𝑦1), 𝑔(𝑦2), … , 𝑔(𝑦𝑜)) into 𝑔 𝑧1 , 𝑔(𝑧2 , … 𝑔 𝑧𝑗 , 𝑔 𝑧𝑗+1 , … , 𝑔(𝑧𝑜))

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1 All comparators must act in the same way for the 𝑔(𝑦𝑗) as they do for the 𝑦𝑗

Bitonic Sort

Bitonic (Merge) Sort is a parallel algorithm for sorting If enough processors are available, bitonic sort breaks the lower bound on sorting for comparison sort algorithm Asymptotic Runtime of 𝑃 𝑜 log2 𝑜 (sequential execution) Very good asymptotic runtime in the parallel case (as we'll see below). Worst = Average = Best case

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Bitonic

Sequence (𝑦1, 𝑦2, … , 𝑦𝑜) is bitonic, if it can be circularly shifted such that it is first monotonically increasing and then monontonically decreasing. (1, 2, 3, 4, 5, 3, 1, 0) (4, 3, 2, 1, 2, 4, 6, 5)

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Bitonic 0-1 Sequences

0𝑗1𝑘0𝑙 1𝑗0𝑘1𝑙

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The Half-Cleaner

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bitonic 1 1 1 1 1 1 1 1 1 1 bitonic bitonic clean

void halfclean(std::vector<T>& a, int lo, int n, boolean dir){ for (int i=lo; i<lo+n/2; i++) compare(a[i],a[i+n/2], dir); }

The Half-Cleaner

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bitonic 1 1 1 1 1 1 bitonic clean bitonic

void halfclean(std::vector<T>& a, int lo, int n, boolean dir){ for (int i=lo; i<lo+n/2; i++) compare(a[i],a[i+n/2], dir); }

Bitonic Split Example

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+

bitonic bitonic bitonic

Lemma

Input: a bitonic sequence of 0s and 1s, then for the output of the half-cleaner it holds that

• Upper and lower half is bitonic
• [One of the two halfs is bitonic clean (only 0s or 1s)]
• Every number in upper half ≤ every number in the lower half

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Proof: All cases

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1 bitonic 1 1 1 1 1 1 bitonic bitonic clean

top bottom top bottom

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1 bitonic 1 1 1 1 bitonic clean bitonic

top bottom top bottom

1 1

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1 bitonic 1 bitonic bitonic clean

top bottom top bottom

1 1

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1 bitonic 1 bitonic bitonic clean

top bottom top bottom

1 1

The four remaining cases (010  101)

51 1 1 bitonic 1 1 1 1 bitonic clean bitonic

top bottom top bottom

1 1 1 1 bitonic 1 1 1 1 bitonic bitonic clean

top bottom top bottom

1 1 1 1 bitonic 1 1 1 bitonic clean bitonic

top bottom top bottom

1 1 1 1 1 1 1 1 bitonic 1 1 1 bitonic clean bitonic

top bottom top bottom

1 1 1 1 1 1

Construction BitonicToSorted

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1 1 1

1 1 1

halfclean halfclean halfclean

1 1 1

half clean half clean half clean half clean

1 1 1

bitonic sorted

Recursive Construction

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halfclean (n)

bitonic ToSorted (n/2) bitonic ToSorted (n/2)

bitonicToSorted(n)

≝

BitonicToSorted sorts a Bitonic Sequence

void bitonicToSorted (std::vector<int>& a, int lo, int n, boolean dir){ if (n>1){ halfClean(a, lo, n, dir); bitonicToSorted(a, lo, n/2, dir); bitonicToSorted(a, lo+m, n-n/2, dir); } }

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halfclean (n)

bitonic ToSorted (n/2) bitonic ToSorted (n/2)

bitonic

1 1 1 1 1 1

sorted

Bi-Merger

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1 1 1 1 1 1 bitonic bitonic sorted sorted 1 1 1 1 1 1 bitonic bitonic sorted reverse sorted

≜ bitonic

Bi-Merger on two sorted sequences acts like a half-cleaner on a bitonic sequence (when one of the sequences is reversed) biMerge half-cleaner

Bi-Merger

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1 1 1 1 1 1 bitonic bitonic sorted sorted Bi-Merger on two sorted sequences acts like a half-cleaner on a bitonic sequence (when one of the sequences is reversed) bimerge

void bimerge(std::vector<T>& a, int lo, int n, boolean dir){ for (int i=0; i<n/2; i++) compare(a[lo+i],a[lo+n-i-1], dir); }

Merger

Merger

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1 1 1

1 1 1

bimerge (n) halfclean (n/2) halfclean (n/2)

1 1 1

half clean half clean half clean half clean

1 1 1

sorted sorted sorted

Merger

Merger

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1 1 1

1 1 1

bimerge (n)

1 1 1 1 1 1

sorted sorted sorted

bitonic ToSorted (n/2) bitonic ToSorted (n/2)

bitonic bitonic

BitonicMerge sorts a Halfsorted Sequence

void bitonicMerge (std::vector<int>& a, int lo, int n, boolean dir){ if (n>1){ int m=n/2; bimerge(a,lo,n,dir); bitonicToSorted(a, lo, m, dir); bitonicToSorted(a, lo+m, n-m, dir); } }

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bitonic ToSorted (n/2) bitonic ToSorted (n/2)

sorted

1 1 1 1 1 1

sorted bimerge sorted

Recursive Construction of a Sorter

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bitonicSort (n/2) bitonic Merge (n)

bitonicSort(n)

≝

bitonicSort (n/2)

private void bitonicSort(std::vector&<T> a, int lo, int n, boolean dir){ if (n>1){ int m=n/2; bitonicSort(a, lo, m, ASCENDING); bitonicSort(a, lo+m, n-m, DESCENDING); bitonicMerge(a, lo, n, dir); } }

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bitonicSort (n/2) bitonic Merge (n) bitonicSort (n/2)

bitonicMerge (8)

Example

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bitonicMerge(4) bitonicMerge(2)

biMerge halfclean halfclean half clean half clean half clean half clean biMerge half clean half clean biMerge half clean half clean half clean half clean half clean half clean

Merger (8)

Example

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Merger(4) Merger (2)

bi-merger half cleaner half cleaner half cleane r half cleane r half cleane r half cleane r bi- merger half cleane r half cleane r bi- merger half cleane r half cleane r half cleane r half cleane r half cleane r half cleane r

Bitonic Merge Sort

How many steps? ෍

𝑗=1 log 𝑜

log 2𝑗 = ෍

𝑗=1 log 𝑜

𝑗 log 2 = log 𝑜 ⋅ (log 𝑜 + 1) 2 = 𝑷(𝐦𝐩𝐡𝟑 𝒐)

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#mergers #steps / merger

Zur Prüfung

• findet statt am 6.8.2018 von 9:00 – 11:00 (2h)
• Inhalt: Datenstrukturen und Algorithmen, C++ Advanced, Parallel

Programming,

• Hilfsmittel: 4 A4 Seiten, handgeschrieben oder min. 11Pt Fontsize.

Kopieren ist erlaubt aber nicht clever. Handgeschriebenes vom Tablet drucken ist auch erlaubt, wenn dabei die Schrift nicht wesentlich verkleinert wird (in Relation zur sonst üblichen Schreibschrift).

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Vorschlag: Q&A vor der Prüfung.

• Schulferien ab 16.Juli -- Termin sollte vorher sein.
• Für Sie: je später desto besser. Vorschlag: +/- 12. Juli.

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Vorbereitung

• Übungen machen / gemacht haben.

[morgen, Freitag 1. Juni, finden Übungen statt. Besprechung Übung 13.]

• Können Sie die Vorlesungsinhalte einem Kollegen (ohne Folien)

erklären?

• Alte Prüfungen stehen auf der Webseite zur Verfügung.
• Die alten Prüfungen von Prof. Widmayer enthalten Material, das nicht behandelt

wurde (geometrische Algorithmen, branch-and-bound)

• Die "neuen" Prüfungen bei mir enthalten Material, das in den Vorlesungen von

Widmayer/Püschel nicht behandelt wurden (insbesondere C++ / Parallel Programming)

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Ausschlüsse

NICHT Prüfungsstoff sind

• Details der längeren Beweise (Laufzeit Algorithmus Blum, Analyse

Ranomisierter Quicksort, Amortisierte Analyse von Move-To-Front, Beweis Theorem Universelles Hashing, Beweis Fibonacci Zahlen mit Erzeugendenfunktion, Beweis Amortisierte Kosten Fibonacci Heap,

• Atomare Register / RMW Operationen / Lock Free Programming
• Hardware Architekturen, Pipelining, Peterson Algorithmus

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Ich bin weiterhin erreichbar unter felix.friedrich@inf.ethz.ch